Consider the given reaction, Identify X and Y:

The reaction sequence can be understood by analyzing each step separately.

In the first step, 2-methylbutanal reacts with hydrogen cyanide ($HCN$) to form a cyanohydrin.

The reaction proceeds through nucleophilic addition, where the cyanide ion ($CN^-$) attacks the electrophilic carbonyl carbon.

Pure $$HCN$$ is a weak acid and ionizes only slightly, so it does not provide a sufficient concentration of $$CN^-$$ ions. Therefore, a base catalyst is required to generate the active nucleophile.

Hence, reagent $$X$$ must be $$NaOH$$, which converts $$HCN$$ into $$CN^-$$. Using an acid such as $$HNO_3$$ would suppress the ionization of $$HCN$$ and inhibit the reaction.

The cyanohydrin formed in the first step contains both a hydroxyl group ($-OH$) and a nitrile group ($-C \equiv N$).

In the second step, treatment with $$LiAlH_4$$ reduces the nitrile group to a primary amine.

The reduction follows the transformation

$$-C \equiv N \longrightarrow -CH_2NH_2.$$

The hydroxyl group remains unchanged during this process.

Therefore, the carbon atom of the nitrile group is retained in the final product, and the product must contain a $$-CH_2NH_2$$ substituent attached to the carbon bearing the hydroxyl group.

Any structure showing the amino group attached directly as $$-NH_2$$ to that carbon is incorrect because it omits the carbon originally present in the cyanide group.

Hence,

• $$X = NaOH$$
• $$Y$$ is the structure containing the $$-CH_2NH_2$$ group.

Therefore, the correct answer is the first option.